Asymptotic behaviour of x(n)

Given the equation: $x^2 + \ln^2x = n$.

Let $x(n)$ - root of the equation and $x(n) > 1$

The problem: find missing terms in following sum: $$x(n) \approx ... + O(\frac{\ln^4n}{n})$$

I've found the first term and get next result:

$$x(n) \approx \sqrt{n} + ... + O(\frac{\ln^4n}{n})$$

How to determine remaining terms of the sum?

Well, let $x=\sqrt{n}-y$, where $y=o(\sqrt n)$. Now: $$(\sqrt{n}-y)^2+\ln^2(\sqrt{n}-y)=n$$ $$-2\sqrt{n}\cdot y + y^2 + \ln^2\sqrt n + \left(\ln(1-\frac{y}{\sqrt n})\right)^2=0$$ Now discard the smaller terms and see what remains: $$-2\sqrt{n}\cdot y + \ln^2\sqrt n = 0$$ $$y = \left.\left({1\over2}\ln n\right)^2\right/2\sqrt{n}$$ (plus some small-o, of course).
• Can you explain why $y^2=o(n)$ has been discarded? It does not look too small. – A.Γ. Oct 16 '15 at 7:07
• If $y=o(\sqrt n)$, then $y^2=o(y\cdot\sqrt n)$. – Ivan Neretin Oct 16 '15 at 7:27
• Yes, but $y=o(\sqrt{n})$ could make $y^2$ much larger than $\ln^2\sqrt{n}$. Thus one cannot discard the former and keep the larger as the answer is presently doing. More work is needed. – Did Oct 16 '15 at 7:29
• If $y^2$ were as large as $\ln^2n$ or larger, then $y\sqrt n$ would be even larger, and wouldn't have anything to cancel out with. I agree, of course, that my answer is but a sketch rather than a proof. – Ivan Neretin Oct 16 '15 at 7:41
• In your example we don't have the reason to believe that the quadratic term is small; $\epsilon$ may be small, but we don't know what $y$ is. In this problem, we do know that $y=o(\sqrt n)$; that was our assumption from the very beginning. Anyway, if you insist, let's just discard the second logarithm term (which is $o(1)$ anyway), solve the quadratic equation for $y$, approximate the solution and get exactly the same result. – Ivan Neretin Oct 16 '15 at 7:53